Trigonometric Equations Question 364
Question: For a man, the angle of elevation of the highest point of the temple situated east of him is $ 60^{o} $ . On walking 240 metres to north, the angle of elevation is reduced to $ 30^{o} $ , then the height of the temple is
[MP PET 2003]
Options:
A) $ 60\sqrt{6}m $
B) $ 60m $
C) $ 50\sqrt{3}m $
D) $ 30\sqrt{6}m $
Show Answer
Answer:
Correct Answer: A
Solution:
- Total distance from temple = $ \sqrt{x^{2}+{{(240)}^{2}}} $ where $ x=\frac{h}{\tan 60^{o}} $ = $ \frac{h}{\sqrt{3}} $ So distance $ =\sqrt{\frac{h^{2}}{3}+{{(240)}^{2}}} $ but $ \frac{h}{\sqrt{\frac{h^{2}}{3}+{{(240)}^{2}}}}=\frac{1}{\sqrt{3}} $
Þ $ \frac{h^{2}}{\frac{h^{2}}{3}+{{(240)}^{2}}}=\frac{1}{3} $ After solving, $ h=60\sqrt{6}m. $
BETA
